# Challenging integral: $\int_0^Z\frac{\alpha^{(1-x^2)}}{1-x^2} dx$

I'd like to find a symbolic form for the following integral:

$$f(\alpha, Z) = \int_0^Z\frac{\alpha^{(1-x^2)}}{1-x^2} dx$$

It is given that $0 \le \alpha \le 1$ and $0 \le Z < 1$.

The following partial solutions are known:

$$f(\alpha, 0) = 0 \\ f(0, Z) = 0 \\ f(1, Z) = \mathrm{arctanh}(Z)$$ Otherwise, the integral appears non-trivial.

I have little hope that a symbolic solution can be derived in terms of elementary functions. But it would still be very useful to write it in terms of non-elementary special functions, such as given on this page.

I mostly need accurate and fast evaluation. Unfortunately, direct integration will be too slow (the function will be used many billions of times during the fitting of data).

Any help would be greatly appreciated!

• You could calculate values $f(\frac{i}{n}, \frac{j}{n})$ for all $i, j \in \{0, 1, 2, .., n \}$ and use interpolation to calculate $f(\alpha , Z)$ in general. – Reactant Aug 9 '15 at 12:56
• Write $\alpha^{1+x^2}=e^{\log(\alpha)(1+x^2)}$. The integral is then in the form of Owen's T function, which is quite well known. en.wikipedia.org/wiki/Owen%27s_T_function – tired Aug 12 '15 at 9:52

You might want to approximate the integral with a series: If we introduce a parameter $b$ to give $$f(\alpha, b, Z) = \int_0^Z\frac{\alpha^{(1-x^2)}}{b-x^2} dx$$ then the inverse Laplace transform with respect to $b \to s$ is $$\mathcal{L}^{-1}_{b \to s}[f(\alpha, b, Z)] = \alpha^{1-x^2}e^{sx^2}$$ then $$\int_0^Z\alpha^{1-x^2}e^{sx^2}dx = \frac{\alpha \sqrt{\pi}}{2}\frac{\mathrm{erf}(Z\sqrt{\log(\alpha)-s})}{\sqrt{\log(\alpha)-s}}$$ and $$f(\alpha, Z) = \frac{\alpha \sqrt{\pi}}{2}\int_0^\infty \frac{\mathrm{erf}(Z\sqrt{\log(\alpha)-s})}{\sqrt{\log(\alpha)-s}} e^{-s} \;ds$$ take the Mellin transform with respect to $Z \to q$ this gives $$\mathcal{M}[f(\alpha,Z)]=\frac{-\alpha}{2q}\int_0^\infty \Gamma\left(\frac{1}{2}+\frac{q}{2}\right)(\log(\alpha)-s)^{-\frac{1}{2}-\frac{q}{2}} e^{-s} \; ds \\ \mathcal{M}[f(\alpha,Z)] = \frac{\Gamma \left(\frac{q+1}{2}\right) \log ^{\frac{1}{2}-\frac{q}{2}}(a) E_{\frac{q+1}{2}}(-\log (a))}{2 q}$$ where $E$ is the exponential integral, then according to the Ramanujan master theorem we can perform the inverse Mellin transform as a sum $$f(a,Z)=\sum_{q=0}^\infty \frac{\Gamma \left(\frac{1-q}{2}\right) \log ^{\frac{1}{2}+\frac{q}{2}}(a) E_{\frac{1-q}{2}}(-\log (a))}{2 \Gamma(1-q)}$$ so we can truncate to approximate the integral, (even $n$ sums vanish) $$f(a,Z) \approx a Z + \frac{1}{2} \sqrt{\pi } \sqrt{\log (a)} E_{\frac{1}{2}}(-\log (a))+\frac{1}{10} Z^5 \Gamma (3,-\log (a))+\frac{1}{3} Z^3 \Gamma (2,-\log (a))+ \cdots$$ where there are incomplete gamma functions present. This approximation seems to hold well for small $\alpha$ and $Z$, and there are some complex answers from the series, but the real part is what you want. There should be existing fast numerical schemes for the special functions in the series.